The answer is no, they are not the same.
The best way to see this is to consider a finite dimensional complex $K$ with an action of a finite group $\rho$. Then in your first definition of equivariant cohomology, $H^*_\rho(K;A)$ vanishes for $*>\dim K$. (This is easy to see, since there are no chains above the dimension of $K$.)
However, if $K$ has a fixed point then the Borel fibration $E\rho\times_\rho K\to B\rho$ admits a section, and it follows that $H^*(B\rho;A)$ injects into $H^*(E\rho\times_\rho K;A)$. So this latter group is often non-trivial in infinitely many degrees.
I would very much like to know the precise relationship between these two equivariant cohomology theories (especially when $A$ is a non-trivial module); I suppose they are related by some spectral sequence.
On a historical note, the definition you attribute to Steenrod is actually due to Eilenberg:
Eilenberg, Samuel, Homology of spaces with operators. I, Trans. Am. Math. Soc. 61, 378-417 (1947). ZBL0034.11001.
Borel cohomology came around a decade later, and Bredon cohomology about a decade later still.
Added later: There is indeed a spectral sequence relating $H^*_\rho(K;A)$ and $H^*(E\rho\times_\rho K;A)$, and it appears in Chapter VII of Brown's book "Cohomology of Groups". Here the assumptions are that $\rho$ is an arbitrary discrete group, $A$ is a $\mathbb{Z}[\rho]$-module, and $K$ is a complex on which $\rho$ acts cellularly.
Brown gives a more algebraic description of the equivariant cohomology of $K$ with coefficients in $A$ which agrees with the Borel cohomology $H^*(E\rho\times_\rho K; A)$ (see Exercise VII.7.3, which treats homology but cohomology is similar). Consider the (ordinary, non-equivariant) cochain complex $C^*(K;A)=\operatorname{Hom}_\mathbb{Z}(C_*(K),A)$. This can be made into a cochain complex of $\mathbb{Z}[\rho]$-modules, via the diagonal action, whereby for a cochain $\varphi$, chain $\sigma$ and group element $g\in \rho$ we have
$$
g\cdot\varphi(\sigma) = g\varphi(g^{-1}\sigma).
$$
Brown then defines the equivariant cohomology of $K$ to be $$
H^*_{B}(K;A):=H^*(\rho; C^*(K;A)),
$$ the group cohomology of $\rho$ with coefficients in the cochain complex $C^*(K;A)$ (the $B$ could stand for Brown or Borel, since they agree). To unravel this definition, take a projective resolution $P_\bullet\to \mathbb{Z}\to 0$ of the trivial $\mathbb{Z}[\rho]$-module, and consider the double complex $D^{*,*}$ given by
$$
D^{p,q}=\operatorname{Hom}_\rho(P^q,C^p(K;A)).
$$
It has differentials of bidegrees $(0,1)$ and $(1,0)$ given by the differentials of $P_\bullet$ and $C^*(K;A)$, respectively. The equivariant cohomology of $K$ is then the cohomology of the total complex $\operatorname{Tot}(D^{*,*})$.
There arise two different spectral sequences converging to $H^*_B(K;A)$, coming from filtering the total complex either horizontally or vertically. One of these gives a spectral sequence whose first page has
$$
E^{p,q}_1 = H^q(\rho ; C^p(K;A))
$$ and with $d_1:E^{p,q}\to E^{p+1,q}$ induced by the differential $C^p(K;A)\to C^{p+1}(K;A)$. Now comes the clever part: when $q=0$ we have
$$
E^{p,0}_1= H^0(\rho ; C^p(K;A)) = C^p(K;A)^\rho = \operatorname{Hom}_\rho(C_p(K),A),
$$
since the $\rho$-fixed points under the diagonal action on the cochains are just the equivairiant cochains. It follows that
$$
E_2^{p,0}=H^p_\rho(K;A),
$$
the classical (Eilenberg) equivariant cohomology defined in your question. The edge homomorphism $H^p_\rho(K;A)\to H^p_B(K;A)$ of this spectral sequence can be used to relate the two types of equivariant cohomology. For instance, they are isomorphic in dimension $p$ if the $E_2^{p-j,j}$ entries are zero for $j>0$. These can be interpreted as certain sheaf cohomology groups of the orbit space $K/\rho$.
